CHEE 311J.S. Parent1 4. Chemical Potential in Mixtures When we add dn moles of a component to n...

CHEE 311 J.S. Parent 1

4. Chemical Potential in Mixtures

When we add dn moles of a component to n moles of itself, we will observe (?) a change in Gibbs energy of:

where i,T,P represents the chemical potential of the pure component.

In dealing with mixtures, we need to ask ourselves what the change in the Gibbs energy of a mixture will be if we add dn moles of a component.

Will the Gibbs energy change follow the above formula? If so, why? What, if any, differences need to be described?

iP,TiP,Tdn)nG(d


Chemical Potential in Mixtures

Because the Gibbs energy is a function of composition, we expect the chemical potential to have a composition dependence as well.

We are looking to derive i as a function of T,P and yi

This requires an expression for the total Gibbs energy of a mixture:

to which the defining equation for i is applied:

10.1

In lecture 5 we derived the dependence of (H, S) on (T,P) for a single component system.

What are the corresponding relationships for mixtures?

nj,P,Tii n

nG

)n,...,n,n,P,T(f

TnSnHnG

k21


Chemical Potential in Perfect Gas Mixtures

The molecular conditions that produce a perfect gas mixture are the same as those of an ideal gas.

The gas must consist of freely moving particles of negligible volume and having negligible forces of interaction

Gibbs remarks “…every gas is a vacuum to every other gas (in the mixture)”

If n moles of an perfect gas mixture occupy V at T, the pressure is:

If nk moles of component k in this mixture were to occupy the same volume V at T:

Which leads to the concept of partial pressure:

(10.20)

VnRT

P

VRTn

P kk

)n,...,2,1k(PyP kk


Properties of Perfect Gas Mixtures

Gibbs Theorem:

A total thermodynamic property of an ideal gas mixture is the sum of the total properties of the individual species, each at the mixture temperature, but at its own partial pressure, pi.

In a mathematical form, a system property M varies according to:

On a molar basis:

= yA * + yB *Gig(T,P)

)p,T(Mn)P,T(nM iigii

ig

)p,T(My)P,T(M iigii

ig

GAig(T,pA) GB

ig(T,pB)


Enthalpy of a Perfect Gas Mixture

For a pure ideal gas at constant temperature:

6.20

6.22

Therefore, the molar enthalpy Hig(T,P) of a pure ideal gas is independent of pressure:

Hig(T,pi) = Hig(T,P)

The total molar enthalpy of an ideal mixture of ideal gases is:

Hig = yi Hiig(T,pi)

or

Hig = yi Hiig(T,P) 10.23

0dH

dPTV

TVdTCdH

T

ig

Pp

0


Entropy of a Perfect Gas Mixture

We have seen that the entropy of a pure ideal gas is a function of T and P according to:

5.13, 6.23

Therefore, the entropy of a pure ideal gas at a partial pressure, pi relative to a total pressure P is for a given temperature:

or

The entropy of an ideal mixture of perfect gases is therefore,

10.25

dPPR

dTT

CdS p

ip

P

igii

igi dP

P

R)P,T(S)p,T(S

iiigii

iiigii

iigii

ig

ylnRy)P,T(Sy

)P/pln(Ry)P,T(Sy

)p,T(Sy)P,T(S

)P/pln(R)P,T(S)p,T(S iigii

igi


Gibbs Energy of an Ideal Gas

The fundamental equation for a closed system gives us the dependence of Gibbs Energy on pressure and temperature:

6.11

For a pure ideal gas, i, at a given temperature, this reduces to:

(constant T)

which integrates to give the Gibbs Energy of this gas as a function of pressure:

or

This is the Gibbs energy of a pure, ideal gas at (T,pi) relative to (T,P). This leads immediately to the Gibbs energy function for an ideal mixture of ideal gases (perfect gas mixture).

SdTVdPdG

dPVdG igi

igi

dPP

RT)T,P(G)T,p(G

pi

P

igii

igi

P

plnRT)T,P(G)T,p(G iig

iiig

i


Gibbs Energy of a Perfect Gas Mixture

Using Gibbs Theorem to define the properties of a perfect gas mixture (slide 5), we have:

In terms of the total Gibbs energy of a system of n moles:

The total Gibbs energy is the sum of: pure component Gibbs energies at reference state

(T,P) Gibbs energy term to account for component mixing

nn

lnnRT)P,T(GnnG ii

igii

ig

iiigii

iiigii

iigii

ig

ylnyRT)P,T(Gy

)P/pln(RTy)P,T(Gy

)p,T(Gy)P,T(G


Chemical Potential in a Perfect Gas Mixture

Calculating the chemical potential for an perfect gas mixture begins with the definition of i:

which applied to our expression for nGig gives:

10.26

jn,P,Ti

igigi n

nG

ioi

iigi

n,P,Ti

ii

igii

igi

ylnRT

ylnRT)P,T(G

nnn

lnnRT)P,T(Gn

j


4. Chemical Potential in Ideal Solutions

We previously developed an expression for the chemical potential of an perfect gas mixture:

Ideality requires freely moving particles of negligible volume and having negligible forces of interaction

If we relax these criteria somewhat we arrive at the more general model of an ideal solution:

Require that all molecules are of the same size and all forces of interaction between molecules (like and unlike) are equal

Clearly, perfect gas mixtures are a special case of the ideal solution model. The relaxed constraints allow us to characterize fluids where interactions are significant, such as liquids and non-ideal gases.


Properties of Ideal Solutions

Molar Volume:

If two liquids of different molar volumes, Vi, are mixed to generate an ideal solution,

Vid = xi Vi 10.82

This results from equivalent forces of interaction (A-A = B-B = A-B). In other words, the components must be virtually identical in every way except for chemical structure.

Enthalpy:

Given that the creation of an ideal solution results in no change in molecular interactions, we do not expect molecular energies to change upon mixing, and

Hid = xi Hi 10.83

where Hi is the entropy of a pure component at the mixture T,P.


Properties of Ideal Solutions

Entropy:

Although mixing of components to generate an ideal solution does not change intermolecular forces (hence U and H are constant), we expect an increase in entropy.

In our development of the entropy of an ideal gas mixture, we applied Gibbs theorem and the pressure dependence of entropy.

A general treatment of the entropy of mixing can be derived from statistical mechanics (See me for references)

The entropy of an ideal solution relative to its pure components is:

Sid = xi Si - R xi ln xi 10.81

where Si is the entropy of pure component i at the mixture T,P.


Chemical Potential of an Ideal Solution

Having defined Hid and Sid, the Gibbs energy and chemical potential immediately follow:

Gid = Hid - T Sid

= xi Hi - T xi Si + RT xi ln xi

= xi Gi + RT xi ln xi 10.80

where Gi represents the pure component Gibbs energy at the mixture T,P

In terms of the total Gibbs energy of a system:

nGid = ni Gi + RT ni ln ni

The chemical potential follows from this equation, leading to

iid = Gi + RT ln xi 10.76

where our reference state is the pure component at T,P


Origin of Raoult’s Law

Ideal Vapour-Liquid Equilibrium (VLE)

The simplest VLE condition exists if a perfect gas mixture is in equilibrium with an ideal liquid solution.

For the vapour phase: iig = Gi

ig + RT ln yi

For the liquid phase: iil = Gi

l + RT ln xi

At equilibrium, the chemical potential of each component must be equal in all phases:

iig = i

id

Using our model equationsGi

ig + RT ln yi = Giil + RT ln xi

orRT ln yi / xi = Gi

il (T,P) - Giig (T,P) (A)

To proceed further, we need to consider the pure component Gibbs energies of both phases.



How do the Gibbs energies of pure component i differ in the vapour and liquid states?

1. Assume Giil is not a strong function of pressure:

Giil (T,P) = Gi

il (T, Pisat)

where Pisat is the vapour pressure of component i.

2. Calculate the change in Gibbs energy when the component as an ideal gas is compressed to the saturation pressure:

Giig (T,P) Gi

ig (T, Pisat)

Integrating from P to Pisat at constant T gives

PRT

VP

G igi

T

igi

satiP

P

igi

sati

igi P

dPRT P)(T, G )P (T, G



The change of Gibbs energy between a pure ideal gas at its saturation pressure and the system pressure, P, is:

At Pisat, the pure component vapour and liquid are in equilibrium.

Therefore, you should convince yourself that:

Giil (T,P) Gi

il (T, Pisat) = Gi

ig (T, Pisat)

and(B)

which is the expression we require to develop Raoult’s Law

Substituting Equation B into Equation A:

RT ln yi / xi = Giil (T,P) - Gi

ig (T,P)

= RT ln Pisat / P

)P/(P ln RT P)(T, G - )P (T, G sati

igi

sati

igi

)P/(P ln RT P)(T, G - )P (T, G sati

igi

sati

ili



Final expression for ideal VLE:

yi / xi = Pisat / P

or

yi P = xi Pisat 12.19

partial pressure mole fraction vapour pressure of i in vapour of i in liquid of pure i

How did we arrive at this expression? Defined an ideal gas mixture

iig = Gi

ig + RT ln yi

Defined an ideal solution i

il = Gil + RT ln xi

Applied the criterion for chemical equilibrium i

ig = iil

= *


Ideal Phase Behaviour: P-xy Diagrams


4. P,T-Flash Calculations

If a stream consists of three components with widely differing volatility, substantial separation can be achieved using a simple flash unit.

Questions often posed:

Given P, T and zi, what are the equilibrium phase compositions?

Given P, T and the overall composition of the system, how much of each phase will we collect?

Feedz1

z2

z3=1-z1-z2

Tf, Pf

P,T

Vapoury1

y2

y3=1-y1-y2

Liquidx1

x2

x3=1-x1-x2


P-T Flash Calculations from a Phase Diagram

For common binary systems, you can often find a phase diagram in the range of conditions needed.

For example, a Pxy diagram for the

furan/CCl4 system at 30C is

illustrated to the right.

Given

T=30C, P= 300 mmHg, z1= 0.5

Determine

x1, x2, y1, y2 and the fraction of the

system that exists as a vapour (V)


Flash Calculations from a Phase Diagram

Similarly, a Txy diagram can be used if available.

Consider the ethanol/toluene system illustrated here at P = 1atm.

Given

T=90C, P= 760 mmHg, z1= 0.25

Determine

x1, x2, y1, y2 and the fraction of thesystem that exists as a liquid (L)

How about:

T=90C, P= 760 mmHg, z1= 0.75?


Phase Rule for Intensive Variables

For a system of phases and N species, the degree of freedom is:F = 2 - + N

# variables that must be specified to fix the intensive state of the system at equilibrium

Phase Rule Variables:The system is characterized by T, P and (N-1) mole fractions for each phase

the masses of the phases are not phase-rule variables, because they do not affect the intensive state of the system

Requires knowledge of 2 + (N-1) variables

Phase Rule Equations:

At equilibrium i = i

= i for all N species

These relations provide (-1)N equations

The difference is F = [2 + (N-1)] - [(-1)N] = 2- +N


Duhem’s Theorem: Extensive Properties SVNA12.2

Duhem’s Theorem: For any closed system of known composition, the equilibrium state is determined when any two independent variables are fixed.

If the system is closed and formed from specified amounts of each species, then we can write:

Equilibrium equations for chemical potentials (-1)N Material balance for each species N We have a total of N equations

The system is characterized by :T, P and (N-1) mole fractions for each phase 2 + (N-1) Masses of each phase Requires knowledge of 2 + N variables

To completely determined requires a knowledge of :[2 + N] - [N] = 2 variables

This is the appropriate “rule” for flash calculation purposes where the overall system composition is specified


Ensuring you have a two-phase system

Duhem’s theorem tells us that if we specify T,P and zi, then we have sufficient information to solve a flash calculation.

However, before proceeding with a flash calc’n, we must be sure that two phases exist at this P,T and the given overall composition: z1, z2, z3

At a given T, the maximum pressure for which two phases exist is the BUBL P, for which V = 0

At a given T, the minimum pressure for which two phases exist is the DEW P, for which L = 0

To ensure that two phases exist at this P, T, zi:

Perform a BUBL P using xi = zi

Perform a DEW P using yi = zi


Ensuring you have a two-phase system

If we revisit our furan /CCl4 system at 30C, we can illustrate this point.

Given

T=30C, P= 300 mmHg, z1= 0.25

Is a flash calculation possible?

BUBLP, x1 = z1 = 0.25

DEWP, y1 = z1 = 0.25

Given

T=30C, P= 300 mmHg, z1= 0.75

Is a flash calculation possible?

BUBLP, x1 = z1 = 0.75

DEWP, y1 = z1 = 0.75


Flash Calculations from Raoult’s Law

Given P,T and zi, calculate the compositions of the vapour and liquid phases and the phase fractions without the use of a phase diagram.

Step 1.

Determine Pisat for each component at T (Antoine’s eq’n, handbook)

Step 2.Ensure that, given the specifications, you have two phases by calculating DEWP and BUBLP at the composition, zi.

Step 3.Write Raoult’s Law for each component:

or(A)

where Ki = Pisat/P is the partition coefficient for component i.

satiii PxPy

iii xKy



Step 4.Write overall and component material balances on a 1 mole basis

Overall:(B)

where L= liquid phase fraction, V= vapour phase fraction.

Component:i=1,2,…,n (C)

(B) into (C) gives

which leads to:(D)

Step 5.Substitute Raoult’s Law (A) into (D) and rearrange:

(E)

mole1FVL

VyLx)1(z iii

Vy)V1(xz iii

V

)V1(xzy ii

i

V

)V1(K

yz

y i

ii

i

)1K(V1

Kzy

i

iii



Step 6:Overall material balance on the vapour phase:

into which (E) is substituted to give the general flash equation:

12.27

where,

zi = overall mole fraction of component iV = vapour phase fraction

Ki = partition coefficient for component i

Step 7:Solution procedures vary, but the simplest is direct trial and error variation of V to satisfy equation 12.27.

Calculate yi’s using equation (E) and xi

’s using equation (A)

1yni

1ii

1)1K(V1

Kzni

1i i

ii

CHEE 311J.S. Parent1 4. Chemical Potential in Mixtures When we add dn moles of a component to n...

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